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Math Help - Ratios of line segments in a triangle

  1. #1
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    Ratios of line segments in a triangle

    I found this problem in one of those foreign Mathematical Olympiad books. The solution is provided, but even then I still can't grasp the steps that lead to it.

    We have a triangle ABC as shown in the figure, with point E selected on side AB in such a way that AE : EB = 1:3, in other words, the ratio of the length of segment AE to the length of segment EB is 1 to 3. Draw the line segment CE.

    We then have a point D selected on side BC so that CD : DB = 1:2. Then draw a line segment from A to D. This gives a point F, which is the intersection of segments AD and CE.

    Solve for (EF/FC) + (AF/FD).
    Attached Thumbnails Attached Thumbnails Ratios of line segments in a triangle-contest-problem-book-2-page-44.jpg  
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  2. #2
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    The solution is a little terse, but goes like this (in bold type):
    1. Draw line DGH parallel to AB. The rest of the solution depends on this very crucial step, but in my pencil and graph paper model, I canít figure out how to get a parallel line drawn at that point, I donít know what to use as a reference length or angle. What I did was build the model in Visio where Iím able to accomplish this and thus proceed with the solution. I use convenient lengths of 8 for AB and 12 for BC, with AC = 14.7508. In my model, line DGH has length 2.6667.
    2. Therefore DG:3a = b:3b and DG = a = EA. Amazingly enough, this is true, but I canít for the life of me figure out or prove why it should be so! The rest of the solution follows, but before I can understand it I need to convince myself of the first two steps. In my model, substituting the lengths in the above equations we obtain 2:6 = 4:12 and 2 = 2 = 2. What I canít grasp is why DG is equal to a.
    3. Therefore EF = FG and AF = FD, so that AF/FD = 1. In my model, EF = FG = 4.5605 and AF = FD = 5.7963, and of course the ratio AF/FD is 1.
    4. Also DH:4a = b:3b, DH = 4a/3 and GH = DH Ė DG = a/3. Iíve got 2.6667:8 = 4:12, 2.6667 = 8/3 and 0.6667 = 2.6667 Ė 2 = 2/3.
    5. Therefore GC = (1/3)EC and EG = (2/3)EC, and since EF = FG, FC = (2/3)EC. In my model, 4.5605 = (1/3) 13.6815 and 9.1210 = (2/3) 13.6815, and since 4.5605 = 4.5605, 9.1210 = (2/3) 13.6815.
    6. Therefore EF/FC = Ĺ. 4.5605 / 9.1210 = Ĺ.
    7. Therefore (EF/FC) + (AF/FD) = Ĺ + 1 = 3/2. (4.5605/9.1210) + (5.7963/5.7963) = Ĺ + 1 = 3/2.
    Attached Thumbnails Attached Thumbnails Ratios of line segments in a triangle-contest-problem-book-2-page-44-solution.jpg  
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